Hooke's Law and the Moduli

Part 2, Part 2: Strain, Elasticity, and the Poroelastic Rock

Learning objectives

State Hooke's law for an isotropic rock: two independent constants price all of elasticity
Assign each modulus its job: K for volume, G for shape, E and nu for the uniaxial-stress experiment, M for the uniaxial-strain one
Convert any modulus pair into all the others and verify the identities, E equal to 2.25 K when K equals G
Bridge to the Rock Physics course: the same moduli set the velocities, and Vp over Vs of 2 means nu of exactly one third

Two Numbers Price Everything

For small strains an isotropic rock is a linear machine: stress in, proportional strain out, and the whole catalog of responses is priced by just two independent constants. Which two is a matter of which experiment you are describing. Squeeze a sample from all sides and volume change is governed by the bulk modulus $K$ ; twist it and shape change is governed by the shear modulus $G$ , the same split the strain tensor made between its trace and its off-diagonal parts. Pull a rod with free sides and its stiffness is Young's modulus $E$ while its sideways bulge is set by Poisson's ratio $\nu$ . Squeeze a layer that cannot bulge sideways, the resting-basin geometry of the next section, and the operative stiffness is the uniaxial modulus $M = K + \tfrac{4}{3}G$ . Any two determine the rest, and fluency in the conversions is a daily-use skill.

Turn the wheel above. Feed it $E$ and $\nu$ , or $K$ and $G$ , and read the full set, with two identities worth pinning. When $K = G$ , Young's modulus is exactly $2.25\,K$ . And when $\nu = 1/4$ , the value our canon rock carries, $K$ and $G$ sit in the tidy ratio $5/3$ . Typical numbers ground the scales: a soft Gulf sand has $E$ near 5 GPa, our Berea-like reservoir sand near 19, a stiff limestone 30 to 55, and $\nu$ runs from 0.20 in clean stiff sandstones toward 0.35 in soft shales.

The Handshake with Rock Physics

These are the same constants the Rock Physics course builds its Part 1 on, because velocities are moduli in disguise: $V_P = \sqrt{(K + \tfrac{4}{3}G)/\rho}$ P=sqrt(K+tfrac43G)/rho and $V_S = \sqrt{G/\rho}$ S=sqrtG/rho, so the ratio $V_P/V_S = \sqrt{M/G}$ depends on the moduli alone. The handshake identity: $V_P/V_S = 2$ means $\nu = 1/3$ exactly, no rock required. That bridge runs both ways and carries a warning this course will repeat in Part 8: seismic and sonic measurements deliver dynamic moduli at tiny strain, while the deformations geomechanics cares about are slower and larger, where rocks respond softer. Static $E$ can be half the dynamic value in weak rock. The conversion between them is calibration work, not a formula, and pretending otherwise is one of the classic ways a stress model goes quietly wrong.

References

Jaeger, J. C., Cook, N. G. W., & Zimmerman, R. W. (2007). Fundamentals of Rock Mechanics (4th ed.). Blackwell.
Fjaer, E., Holt, R. M., Horsrud, P., Raaen, A. M., & Risnes, R. (2008). Petroleum Related Rock Mechanics (2nd ed.). Elsevier.
Mavko, G., Mukerji, T., & Dvorkin, J. (2009). The Rock Physics Handbook (2nd ed.). Cambridge University Press.